The auxiliary equation of y"" - 4y" - 4y' + 16y = 0 is r34r²4r + 16 = 0. The auxiliary equation has a root in the interval [-0.75,3.75]. Let f(r) = r³ − 4r² − 4r + 16 so that the roots I of the auxiliary equation can be determined by solving ƒ(r) = 0. Then, applying four
The auxiliary equation of y"" - 4y" - 4y' + 16y = 0 is r34r²4r + 16 = 0. The auxiliary equation has a root in the interval [-0.75,3.75]. Let f(r) = r³ − 4r² − 4r + 16 so that the roots I of the auxiliary equation can be determined by solving ƒ(r) = 0. Then, applying four
Calculus: Early Transcendentals
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![The auxiliary
equation of
y"" – 4y" – 4y' + 16y = 0
is
r³ – 4r² – 4r + 16 = 0.
The auxiliary
equation has a root
in the interval
[-0.75, 3.75].
Let
f(r) = r³ - 4r² - 4r + 16
so that the roots
"
of the auxiliary
quation can be
determined by solving
f(r) = 0.
Then, applying four
iterations of the
Bisection Method to
f(r), using the
55](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F127cb3e0-a29c-4a03-97bb-e0ced4c9c118%2Fc595d206-8627-4c74-8541-5c297e672987%2Fp5d7ugf_processed.jpeg&w=3840&q=75)
Transcribed Image Text:The auxiliary
equation of
y"" – 4y" – 4y' + 16y = 0
is
r³ – 4r² – 4r + 16 = 0.
The auxiliary
equation has a root
in the interval
[-0.75, 3.75].
Let
f(r) = r³ - 4r² - 4r + 16
so that the roots
"
of the auxiliary
quation can be
determined by solving
f(r) = 0.
Then, applying four
iterations of the
Bisection Method to
f(r), using the
55
![Then, applying four
iterations of the
Bisection Method to
f(r), using the
initial
approximations
r₁ = -0.75 and
r₂ = 3.75, gives the
following
approximations for
the root in
[-0.75,3.75]:
(1)
13
(2)
13
(3)
13
1;(1)
=
=
=
=
55](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F127cb3e0-a29c-4a03-97bb-e0ced4c9c118%2Fc595d206-8627-4c74-8541-5c297e672987%2Feate1mo_processed.jpeg&w=3840&q=75)
Transcribed Image Text:Then, applying four
iterations of the
Bisection Method to
f(r), using the
initial
approximations
r₁ = -0.75 and
r₂ = 3.75, gives the
following
approximations for
the root in
[-0.75,3.75]:
(1)
13
(2)
13
(3)
13
1;(1)
=
=
=
=
55
Expert Solution
Step 1
Given;
f(r)=r3-4r2-4r+16 , [-0.75,3.75].
Step by step
Solved in 3 steps with 2 images
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